Force Homework: Mechanics Theorem - Find the Force in terms of Theta

[lifeonfire]

Homework Statement

Let a mass move in an orbit given by r = a*(theta)
(a) If theta is a linear function of time t, what is the Force in terms of theta?
(b) How should theta depend on t so that the force is central?

Homework Equations

The Attempt at a Solution

Given that theta is a "linear function of time". I can write: theta = A t + B
Thus, r = a*(At +B)
and, v = dr/dt = a*(t)
and a = dv/dt = a

Therefore Force = m*a ? Where m = mass.
The force doesn't depend on theta?
Also, I have no clue what the second question is asking for?

 

[CompuChip]

What do you get when you differentiate a*A*t? It's not a*t!
You should be more careful anyway, the "r" here is just the radial component (i.e. the distance to the origin) in polar coordinates.
Of course, these coordinates are defined by
x(t) = r \cos\theta, y(t) = r \sin\theta,
where the usual notation \vec r(t) = (x(t), y(t)) is causing some confusion here.

Once you solved a, I suggest repeating it for a more general \theta(t). If you equate the force you then calculate to the standard expression for central force, you will get a differential equation for \theta.

 

[collinsmark]

Hello lifeonfire,

I had posted something yesterday, but I quickly deleted it after recognizing a misleading mistake I made. I think I have it right this time.

Given that theta is a "linear function of time". I can write: theta = A t + B
Thus, r = a*(At +B)

Okay, so far so good. Here r is the magnitude of the position vector r.

and, v = dr/dt = a*(t)

This is where things seem to be going wrong. Given the way the problem was phrased, it seems r is a vector. It has a magnitude r and a direction that points θ radians away from the x-axis.

Let's take a step back for a moment. Imagine that we have such a vector r and we want to "wiggle" both r and θ by small (infinitesimal) amounts, and calculate the resulting differential vector (the difference between the vector r before and after the wiggling). In polar coordinates, this vector is

{\bold d} {\bold r} = dr \hat r + rd \theta \hat \theta

dr \hat r is the radial component extending away from the center. Perpendicular to that is tangential component rd \theta \hat \theta. Note that the tangential component contains an r in it. That's because as we wiggle θ by a given amount, the effect that it has on the differential vector is proportional to the radius.

Now divide everything by dt and you have the velocity.

\dot {\bold r} = \dot r \hat r + r \dot \theta \hat \theta

Differentiate that with respect to time, and you have the acceleration vector. (Don't forget to use the chain rule on the \hat \theta component.)

Now you can make your substitutions, differentiating terms as appropriate.

Also, I have no clue what the second question is asking for?

A central force is a conservative force in the direction of \hat r. Gravity and the coulomb forces are examples. But what it really boils down to is that a central force will be a function of r and will be completely in the \hat r direction.

That last part is pretty important. Essentially what the second part of this problem is asking is "how does θ have to vary with time such that the \hat \theta component of the force is zero?"

I'm going to give you a big hint. When you set the \hat \theta component of the acceleration vector equal to zero, and solve for θ, you end up solving for a second-order, nonlinear, ordinary differential equation. Yuck. My hint is as follows: You've already done half the work in your previous steps! Saying that the \hat \theta component of the acceleration must be equal to zero is the same thing as saying that the \hat \theta component of the velocity must be equal to a constant. An arbitrary constant. By recognizing this, you'll end up only needing to solve a 1st order diffy-Q instead of a 2nd order.

 

[lifeonfire]

Thanks :)